Show in Figure 1 is the water flow system. The velocity head difference is . Volume flow rate is . The heights are and , respectively. The power output from the motor is , . The energy added by the pump is . The energy loss between point 1 and 2 is . Calculate 1) power delivered by the fluid to the motor in kW. ___________kW
Show in Figure 1 is the water flow system. The velocity head difference is . Volume flow rate is . The heights are and , respectively. The power output from the motor is , . The energy added by the pump is . The energy loss between point 1 and 2 is . Calculate 1) power delivered by the fluid to the motor in kW. ___________kW
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Show in Figure 1 is the water flow system. The velocity head difference is . Volume flow rate is . The heights are and , respectively. The power output from the motor is , . The energy added by the pump is . The energy loss between point 1 and 2 is .
Calculate 1) power delivered by the fluid to the motor in kW. ___________kW
![### Water Flow System Overview
#### Diagram Description
The given diagram displays a water flow system, which includes a motor, a pump, and various points in the system where measurements are taken (labeled as points 1 and 2). The system has the following notable components:
- **Motor and Pump:**
- The motor drives the pump that moves water through the system.
- The pump is situated near the valve where the flow is regulated.
- **Piping and Flow Points:**
- The piping system includes height changes denoted by \(h1\) and \(h2\).
- Heights are specified as \(h1 = 1.0 \, m\) and \(h2 = 2.0 \, m\).
#### Known Parameters
- **Velocity Head Difference**:
\[
\frac{v_1^2 - v_2^2}{2g} = 2.0 \, m
\]
- **Volume Flow Rate**:
\[
Q = 0.5 \, m^3/s
\]
- **Heights**:
\[
h1 = 1.0 \, m, \quad h2 = 2.0 \, m
\]
- **Motor Power Output**:
\[
P_0 = 7.848 \, kW
\]
- **Motor Efficiency**:
\[
e_{M, motor} = 80\%
\]
- **Energy Added by the Pump**:
\[
h_A = 3.0 \, m
\]
- **Energy Loss Between Points 1 and 2**:
\[
h_L = 0.2 \, m
\]
#### Problem Statement
Calculate the power delivered by the fluid to the motor \( P_R \) in \( kW \).
### Solution Steps
To solve for the power delivered by the fluid to the motor, we follow these steps:
1. **Convert Motor Efficiency to Decimal:**
\[
e_{M, motor} = \frac{80}{100} = 0.8
\]
2. **Find Motor Power Received by the Water:**
Using the efficiency formula:
\[
P_R = \frac{P_0}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0daff6b4-7b3b-4798-aa81-781ef422795b%2F9725e33a-65a6-4003-a260-8ea05e017ef3%2Ffhpyehk_processed.png&w=3840&q=75)
Transcribed Image Text:### Water Flow System Overview
#### Diagram Description
The given diagram displays a water flow system, which includes a motor, a pump, and various points in the system where measurements are taken (labeled as points 1 and 2). The system has the following notable components:
- **Motor and Pump:**
- The motor drives the pump that moves water through the system.
- The pump is situated near the valve where the flow is regulated.
- **Piping and Flow Points:**
- The piping system includes height changes denoted by \(h1\) and \(h2\).
- Heights are specified as \(h1 = 1.0 \, m\) and \(h2 = 2.0 \, m\).
#### Known Parameters
- **Velocity Head Difference**:
\[
\frac{v_1^2 - v_2^2}{2g} = 2.0 \, m
\]
- **Volume Flow Rate**:
\[
Q = 0.5 \, m^3/s
\]
- **Heights**:
\[
h1 = 1.0 \, m, \quad h2 = 2.0 \, m
\]
- **Motor Power Output**:
\[
P_0 = 7.848 \, kW
\]
- **Motor Efficiency**:
\[
e_{M, motor} = 80\%
\]
- **Energy Added by the Pump**:
\[
h_A = 3.0 \, m
\]
- **Energy Loss Between Points 1 and 2**:
\[
h_L = 0.2 \, m
\]
#### Problem Statement
Calculate the power delivered by the fluid to the motor \( P_R \) in \( kW \).
### Solution Steps
To solve for the power delivered by the fluid to the motor, we follow these steps:
1. **Convert Motor Efficiency to Decimal:**
\[
e_{M, motor} = \frac{80}{100} = 0.8
\]
2. **Find Motor Power Received by the Water:**
Using the efficiency formula:
\[
P_R = \frac{P_0}{
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